Chapter 2 · Vectors CHAPTER 3 · PARTICLE EQUILIBRIUM →

Cartesian Vectors

Cartesian vector form expresses a vector using components along orthogonal axes. This representation replaces most geometric construction with component-wise algebra, which scales efficiently when many vectors are involved.

Cartesian Vectors

Graphical vector addition methods, such as the parallelogram law or the tip-to-tail method, are useful for visualizing a small number of vectors. As the number of vectors increases, these methods become cumbersome. For example, combining three forces requires applying the parallelogram law twice—first to obtain the resultant of two forces (R1 in the figure below), and then to combine that resultant with the third (R2).

Graphical addition of three forces using the parallelogram law

In Cartesian form, each vector is represented by its components along the \(x\)-, \(y\)-, and \(z\)-axes (\(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\)). Vector addition is performed by summing the corresponding scalar components along each axis.

Cartesian Vector Addition (Component Form)

\[ R_x = \sum F_{x} \] \[ R_y = \sum F_{y} \] \[ R_z = \sum F_{z} \]

Equivalent Vector Form

\[ \vec{R} = \sum \vec{F}_i \]

Textbook ordering. Some statics textbooks introduce Cartesian components only in two dimensions and defer three-dimensional components until equilibrium is discussed. This page presents both 2D and 3D Cartesian vector representations together as a mathematical reference, independent of their later application.

2D Cartesian Components

In two dimensions, a vector is expressed using unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) aligned with the \(x\)- and \(y\)-axes.

2D Cartesian Vector

\[ \mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} \]

Computing 2D Components Using Trigonometry

When a vector has known magnitude \(A\) and a defined reference angle \(\theta\), its Cartesian components are obtained from right-triangle geometry. The component adjacent to \(\theta\) uses cosine, and the component opposite \(\theta\) uses sine.

Vector components from magnitude and reference angle

\[ A_x = A\cos\theta \]

\[ A_y = A\sin\theta \]

Angle–component rule. Cosine gives the component adjacent to the reference angle, and sine gives the component opposite the angle. The choice depends on the angle definition, not the axes.

Magnitude and Angle in 2D

The magnitude of \(\mathbf{A}\) is obtained from the Pythagorean theorem and the direction angle \(\theta\) (measured from the positive \(x\)-axis) can be computed from: :

Magnitude and Direction Angle (2D)

Vector A with components Ax, Ay and direction angle theta

\[ \|\mathbf{A}\| = \sqrt{A_x^2 + A_y^2} \]

\[ \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right) \]

Angle determination. Draw the \(F_x\) component first along the \(x\)-axis, then draw the \(F_y\) component tip-to-tail. The direction of the resultant vector constructed this way uniquely defines the correct angle with respect to the reference axis, without requiring a separate quadrant check.

Worked Example — Cartesian Vector Notation

Example Cartesian Vectors

Problem

Two forces act at a point in the \(x\)-\(y\) plane.

\(F_1 = 30\,\text{kN}\) directed \(30^\circ\) from the negative y axis.
\(F_2 = 26\,\text{kN}\) directed along a \(5\!-\!12\!-\!13\) direction triangle in the negative x and positive y directions.

Determine the resultant \(\mathbf{R}\) (in Cartesian form), its magnitude, and the angle of \(\mathbf{R}\) measured from the \(+y\)-axis.

Two-force system for Cartesian vector notation example

Resolve each force into \(x\) and \(y\) components

Write each force in Cartesian form \(\mathbf{F}=F_x\,\mathbf{i}+F_y\,\mathbf{j}\).

Force \(F_1\)

When the vector \(F_1\) is resolved into components, it points in the negative \(x\)- and negative \(y\)-directions. Therefore, both the \(x\)- and \(y\)-components are assigned negative signs.

\[ F_{1x} = -30\sin(30^\circ) = -15.0\ \text{kN} \] \[ F_{1y} = -30\cos(30^\circ) = -25.9\ \text{kN} \] \[ \mathbf{F}_1 = (-15.0\,\mathbf{i}-25.9\,\mathbf{j})\ \text{kN} \]

Force \(F_2\) (5–12–13 direction)

The direction triangle represents how the force is oriented. The horizontal side divided by the hypotenuse gives the fraction of the force acting in the \(x\)-direction, and the vertical side divided by the hypotenuse gives the fraction acting in the \(y\)-direction. Multiplying these fractions by the force magnitude gives the components. The signs follow from the direction of the force.

\[ F_{2x} = -26\left(\frac{5}{13}\right) = -10.0\ \text{kN} \] \[ F_{2y} = 26\left(\frac{12}{13}\right) = 24.0\ \text{kN} \] \[ \mathbf{F}_2 = (-10.0\,\mathbf{i}+24.0\,\mathbf{j})\ \text{kN} \]

Sum components to obtain the resultant \(\mathbf{R}\)

Add components along each axis:

\[ R_x = F_{1x}+F_{2x} = -15.0-10.0 = -25.0\ \text{kN} \] \[ R_y = F_{1y}+F_{2y} = -25.9+24.0 = -1.9\ \text{kN} \] \[ \mathbf{R} = (-25.0\,\mathbf{i}-1.9\,\mathbf{j})\ \text{kN} \]

Compute the magnitude of the resultant

\[ |\mathbf{R}|=\sqrt{R_x^2+R_y^2} =\sqrt{(-25.0)^2+(-1.9)^2} =25.1\ \text{kN} \]

Determine the angle from the \(+y\)-axis

Since the angle is referenced from \(+y\), use a right-triangle relationship based on the resultant components and add 90\(^\circ\). The acute angle \(\theta\) from the \(+y\)-axis satisfies \(\tan\theta = |R_x|/|R_y|\).

Two-force system for Cartesian vector notation finding theta

\[ \theta = \tan^{-1}\!\left(\frac{|R_x|}{|R_y|}\right) = \tan^{-1}\!\left(\frac{1.9}{25.0}\right) = 4.3^\circ + 90^\circ = 94.3^\circ \]

Final Answers:

\(\mathbf{R} = (-25.0\,\mathbf{i}-1.9\,\mathbf{j})\ \text{kN}\)
\(|\mathbf{R}| = 25.1\ \text{kN}\)
\(\theta = 94.3^\circ\) from the \(+y\)-axis (toward \(-x\))

Two-force system for Cartesian vector notation magnitude and theta

3D Cartesian Components

In three dimensions, a vector is expressed using \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), aligned with the \(x\)-, \(y\)-, and \(z\)-axes.

3D Cartesian Vector

\[ \mathbf{A}=A_x\mathbf{i}+A_y\mathbf{j}+A_z\mathbf{k} \]

Magnitude, Direction Angles, and Direction Cosines in 3D

The magnitude of a 3D vector and the direction angles \(\alpha\), \(\beta\), and \(\gamma\), the angles between \(\mathbf{A}\) and the positive \(x\)-, \(y\)-, and \(z\)-axes, respectively, are given by:

Magnitude and Direction Angles (3D)

3D vector showing magnitude and direction angles alpha beta gamma

\[ \|\mathbf{A}\|=\sqrt{A_x^2+A_y^2+A_z^2} \]

\[ \cos\alpha=\frac{A_x}{\|\mathbf{A}\|},\quad \cos\beta=\frac{A_y}{\|\mathbf{A}\|},\quad \cos\gamma=\frac{A_z}{\|\mathbf{A}\|} \]

The direction cosines are the normalized components of \(\mathbf{A}\). They describe how the vector is oriented relative to the coordinate axes and are obtained by dividing each component by the vector magnitude.

Unit Direction Vector

\[ \hat{\mathbf{u}}_A=\frac{\mathbf{A}}{\|\mathbf{A}\|} =\left(\frac{A_x}{\|\mathbf{A}\|}\right)\mathbf{i} +\left(\frac{A_y}{\|\mathbf{A}\|}\right)\mathbf{j} +\left(\frac{A_z}{\|\mathbf{A}\|}\right)\mathbf{k} \]

A useful consistency check is that the direction cosines satisfy:

\[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \]

Consistency check. Since \(\hat{\mathbf{u}}_A\) has unit magnitude, the sum of the squares of the direction cosines must equal one.

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